MARKING SCHEME SET 55/1/RU Q. No. Expected Answer ......MARKING SCHEME SET 55/1/RU Q. No. Expected...

of 16 Final draft 17/03/15 04:30p.m.

MARKING SCHEME

SET 55/1/RU

Q. No. Expected Answer / Value Points Marks Total

Marks

Section A

Set1, Q1

Set2,Q5

Set3,Q4

Self inductance of the coil is numerically equal to magnetic flux linked with it

when unit current flows through it. / Self inductance is numerically equal to

induced emf in the coil when rate of change of current is unity.

Unit- Henry or / volt−second/ ampere / weber ampere-1

½

½

1

Set1, Q2

Set 2,Q3

Set 3,Q1

Scattering of the blue colour is maximum due to its shorter wavelength / As

per Rayleigh scattering law, the amount of scattering varies inversely with the

fourth power of wavelength.

1

1

Set1, Q3

Set 2,Q4

Set 3,Q5

T1

Since slope( ) of T1 is greater / Resistance of the wire at T1 is

lower.

½

½

1

Set1, Q4

Set 2,Q2

Set 3,Q3

Point to Point communication mode 1

1

Set1, Q5

Set 2,Q1

Set 3,Q2

Due to conservative nature of electric field / These lines start from the

positive charges and terminate at the negative charges.

Alternatively,

There are two kinds of electric charges (positive and negative) (which acts as

the „source‟ and „sink‟ for the electric field lines.)

1

1

Section B

Set1, Q6

Set 2,Q8

Set 3,Q10

But

Hence ,

4 i.e.

½

½

½

½

Formula for Energy ½

Formula for de-Broglieg wavelength ½

Calculation ½

Effect on wavelength ½

of 16 Final draft 17/03/15 04:30p.m.

Alternatively

Velocity of electron in nth

state

½

½

½

½

2

2

Set1, Q7

Set 2,Q6

Set 3,Q9

1. Size of the antenna or aerial or

2. Increase in effective power radiated by an Antenna ( OR

Power radiated )

3. To minimize mixing of signals from different transmitters

(Any two )

1 + 1

2

Set1, Q8

Set 2,Q9

Set 3,Q7

According to Kirchoff‟s Junction law at B

(As I2=0 (given))

Applying second law to loop AFEB

From A to D along AFD

[Alternatively, if the student determine value of VAD by finding the value of

R, award full marks.]

[ Note: If the student just writes Kirchoff‟s rules, award ½ mark]

½

½

½

½

2

Any two Factors 1 + 1

Labeling of current in different branches of the circuit ½

Calculation 1

Result ½

of 16 Final draft 17/03/15 04:30p.m.

Set1, Q9

Set 2,Q10

Set 3,Q8

OR

For objective lens,

Alternatively,

Angular size of the object=

Angular size of image=

Height of image=

½

½

½

½

½

½

½

½

½

½

1

2

2

Set1, Q10

Set 2,Q7

Set 3,Q6

=

For shortest wavelength in Balmer series

½

Formula for magnification ½

Substitution and Calculation 1

Result ½

Formula for magnification ½

Calculation & Result ½

Angular magnification ½

Height of image ½

Formula ½

Substitution of correct value in formula ½

Value of ½

Region of wavelength ½

of 16 Final draft 17/03/15 04:30p.m.

=

[Note: Since the value of R is not given, award full marks to the candidate if

he writes ]

It will lie in Ultra Violet region

(Give ½ mark if the student just writes, visible region)

½

½

½

2

Section C

Set1, Q11

Set 2,Q18

Set 3,Q15

Net Capacitance , =

=

C =

Net Charge on Capacitors

q = CV

= C

= C

= 600 (0.6 mC)

=

= 20 V

Energy stored in capacitoracross =

½

½

½

½

½

½

3

Set1, Q12

Set 2,Q19

Set 3,Q16

There being a random distribution, in the velocities of the charge carriers,

their average velocity can be taken to be zero.

We have, F = ma = e FE (FE =electric field)

If τ is the average time between collisions (called „relaxation time‟)

½

½

½

Formula for net capacitance and its calculation ½ + ½

Calculation for net charge ½

Formula and calculation for P.d ½

Formula and calculation for energy stored ½ + ½

+

½

Region of wavelength ½

Derivation of the Relation 2

Effect on drift velocity 1

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Now , FE = For given E, the field becomes when the length is

made 3 times. Hence,

[ Note: If explained by any other appropriate method award 1 mark for the

explanation]

½

½

½

3

Set1, Q13

Set 2,Q20

Set 3,Q17

Let unpolarized light be incident on a polaroid; its electric vectors, oscillating

in a direction perpendicular to that of the alignment of the molecules in the

polaroid, are able to pass through it while the component of light along the

aligned molecules gets blocked. Hence the light gets linearly polarised.

[ Note : If student gives labelled diagram, award full marks.]

I1 will remain unaffected whereas I2 will decrease from maximum (=Io/2) to

zero of the incident light.

I2 = I1 cos2 θ / I2 = (Io / 2) cos

2 θ

1

1

1

3

Set1, Q14

Set 2,Q21

Set 3,Q18

The ratio of amplitude of modulating signal (Em) and amplitude of carrier

wave (EC) is called modulating index.

[ Note: Also accept if only the formula ( ) is given]

To avoid /minimize distortion:

Given:

fc=1.5 M Hz

fm=10 kHz =0.01 MHz

USB frequency =fc+ fm

=(1.5+0.01)MHz

=1.51 MHz

LSB frequency= fc-fm

=(1.5-0.01)MHz

=1.49 MHz

1

½

½

½

½

3

Explanation of Polarization through polarizer 1

Variation in I1 and I2 1

Relation between I1 and I2 1

Definition of Modulation index 1

Reason ½

Calculation of USB and LSB ½ +½

Amplitude of AM ½

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Set1, Q15

Set 2,Q22

Set 3,Q11

Trajectory will be a helix

Explanation/Reason

The particle will describe a circle in the y-z plane, due to the component, ,

of its velocity. It also moves along the x-axis (parallel to the field), due to the

component of its velocity. Hence its trajectory would be helical.

Distance moved along the magnetic field in one rotation

x= υx × T

1

1

½

½

3

Set1, Q16

Set 2,Q14

Set 3,Q12

(a) In LCR circuit

Now

and Ω

=1

½

½

½

½

Trajectory of particle 1

Reason /explanation 1

Expression for distance travelled 1

(a) Value of phase difference 2

(b) Value of additional Capacitance 1

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(b) Power Factor

When power factor=1, we have XL=XC

This gives

We , therefore, need to add a capacitor of capacitance (10-2)µF=8µF in

parallel with the given capacitor.

Alternatively,

Let addition capacitance C1 be connected

½

½

½

½

3

Set1, Q17

Set 2,Q15

Set 3,Q13

Generalized form of Ampere Circuital law:

It signifies that the source of magnetic field is not just due to the conduction

electric current(ic) due to flow of charge but also due to the time rate of

change of electric field called displacement current .

During charging and discharging of a capacitor the electric field between the

plates will change so there will be a change of electric flux (displacement

current) between the plates.

1

1

1

3

Set1, Q18

Set 2,Q16

Set 3,Q14

In ∆ ABC

1

½

Generalized form of Ampere‟s Circuital law 1

Significance 1

Explanation 1

Labelled Diagram 1

Verification of Snell‟s law 2

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In ∆ CEA

or ------- It is Snell‟s law.

½

½

½

3

Set1, Q19

Set 2,Q17

Set 3,Q21

Gate P : AND

Gate Q: NOT

Truth table

Input

X

Y A B

0 0 0 1

0 1 0 1

1 0 0 1

1 1 1 0

Equivalent Gate: NAND

½

½

1

½

½

3

Set1, Q20

Set 2,Q11

Set 3,Q22

Name of Gates P and Q ½ + ½

Truth Table 1

Equivalent Gate ½

Logic symbol of equivalent Gate ½

Labeled Circuit diagram 1

Working of Amplifier 1

Expression for voltage gain ½

Expression for current gain ½

of 16 Final draft 17/03/15 04:30p.m.

The input signal, connected between the emitter and base, along with the

forward bias, causes corresponding large changes in output voltage across R.

Current gain

Voltage gain

1

1

½

½

3

Set1, Q21

Set 2,Q12

Set 3,Q19

(a) Characteristic properties of Nuclear force

i) Short range force

ii) Saturation forces

iii) Very Strong force

iv) Charge independent

(Any Three)

(b)

Conclusions

i) Nuclear force is attractive for distance larger than ro

ii) Nuclear force is repulsive if two nucleons are separated by

distance less than ro

iii) Nuclear force decreases very rapidly for r > ro

iv) Potential energy is minimum at / Equilibrium position

(any two)

½

+½+½

½

½ + ½

3

Three characteristic properties ½ + ½ + ½

Graph for potential energy ½

Two conclusions ½ + ½

of 16 Final draft 17/03/15 04:30p.m.

Set1, Q22

Set 2,Q13

Set 3,Q20

(a) 1. There is no emission of photoelectrons i.e. no current if the frequency

of the incident radiation is below a certain minimum value however large may

be the intensity of the light.

2 The current varies directly with the intensity of the incident radiation.

3.The current becomes zero at a certain value of negative potential, applied at

the anode , this is known as stopping potential.

4. The value of stopping potential increases with the increase in the frequency

of the incident radiation.

5.Maximum kinetic energy of the photo electrons does not depend upon

intensity of light..

6.Maximum kinetic energy of photoelectron increases with the frequency of

the incident radiation.

7.The process of photoelectric emission is instantaneous.

(Any three)

(b) It fails to explain why

1.The photo electric emmission is instantaneous.

2.There exists a threshold frequency for a given metal.

3.The maximim KE of photoelectrons is independent of the intensity of

incident radiation.

OR

(a)

i) The energy of a photon is

ii)Each photon is completely absorbed by a single electron.

(b)

Alternatively, or

or

(Any one)

i. When Incident frequency < Threshold frequency, there will be no

emission of electrons. Hence , frequency of incident radiation should

be greater than threshold frequency.

ii. At

Vo is called stopping potential.

½ +

½ +

½

1 ½

½ + ½

1

½

½

3

3

(a) Three experimental observations ½ +½+½

(b) Failure of wave theory 1 ½

(a) Two properties of photon ½ + ½

(b) Eienstein equation 1

Explanation of threshold frequency ½

Stopping potential ½

of 16 Final draft 17/03/15 04:30p.m.

Section D

Set1, Q23

Set 2,Q23

Set 3,Q23

(i) voltage = 220 V

frequency = 50 Hz

(ii) a) It can be stepped up / stepped down

b)It can be converted into d.c

c)Line losses can be minimised

(any one )

(iii) No

(iv) Helping / Brave / Kind / Knowledge about AC or DC / Knowledge

about insulator & conductors/ Awareness about safety precautions.

(any two)

½

½

½

½

1+1

3

Section E

Set1, Q24

Set 2,Q25

Set 3,Q26

a) Total number of electric lines of force passing perpendicular through a

given surface.

Unit – newton m2 / coulomb (or V-m)

According to Gauss theorem, the electric flux through a closed surface

depends only on the net charge enclosed by the surface and not upon the

shape or size of the surface.

For any closed arbitrary slope of the surface enclosing a charge the

outward flux is the same as that due to a spherical Gaussian surface

enclosing the same charge.

Justification: This is due to the fact

(i) electric field is radial and

(ii) the electric field

b)

According to Gauss theorem ,

( charge inside the shell is zero.)

, But dS 0

1

½

½

1

1 + 1

5

Value of voltage and frequency in India ½ +½

Reason of A.C being used more ½

Use of transformer with D.C ½

Two qualities of Anil 1+ 1

(a) Definition of electric flux and unit 1 + ½

Justification 1½

(b) Proof 1+1

of 16 Final draft 17/03/15 04:30p.m.

OR

(a)

Energy Density

But and

(b) Energy before connecting

After connecting

Common potential =

=

Energy Stored

= U

½

½

½

½

½

½

½

½

½

½

5

Set1, Q25

Set 2,Q26

Set 3,Q24

(a) Derivation for energy stored 2

Derivation for energy density 1

(b) Required Proof 2

Labelled diagram 1

Principle and working ½ + 1

Function of radial magnetic field and soft iron core ½ + ½

Current sensitivity ½

Voltage sensitivity ½

Explanation ½

of 16 Final draft 17/03/15 04:30p.m.

Principle : “Whenever a current carrying coil is placed in magnetic field, it

experiences a deflecting torque.”

Working: When current is passed through a coil , free to rotate in a magnetic

field , a deflecting torque (=NiABsin ) act on it. The coil starts to rotate . The

rotation of coil is opposed, by spring Sp by providing a restoring torque

(=K ). When the two torque becomes equal , coil comes to rest.

, Hence

Functions of (1) Radial field ; It keeps magnetic field lines normal

to the area vector of the coil

(2) Soft iron core; It increases the strength of magnetic field.

Current sensitivity = deflection per unit current /

Voltage sensitivity : deflection per unit voltage /

If N⟶ 2N , then by increasing number of turns, current sensitivity increases

but voltage sensitivity remains same because resistance increases

proportionally.

OR

1

½

½

½

½

½

½

½

½

5

(a) Expression for vector form of Biot-Savart law 1

Expression for magnetic field due to loop 3

(b) Biot-Savart law and Ampere‟s Circuital law 1

of 16 Final draft 17/03/15 04:30p.m.

(a) Biot-Savart law in vector form

=

Magnetic field on the axis of a circular current loop

The net magnetic field is along the x-axis only.

Net contribution along X-axis

1

½

½

½

½

½

½

of 16 Final draft 17/03/15 04:30p.m.

(b) Biot-Savart law can be expressed as Ampere‟s circuital law by

considering the surface to be made up a large number of loops. The sum of

the tangential components of the magnetic field multiplied by the length of all

such elements, gives the result

Alternatively ,

Ampere Circuital law and Biot-Savart law , both relate the magnetic field and

the current , and both express the same physical consequences of a steady

current.

1

5

Set1, Q26

Set 2,Q24

Set 3,Q25

(a) The resultant displacement will be

The amplitude of the resultant displacement is

∴ Intensity

If ….. the intensity will be maximum. i.e

where n = 1, 2, 3 ..

Hence interference will be constructive.

If ….., the intensity will be zero, i.e

where n=1, 2, 3 ..

Hence interference will be destructive.

(b)(i)Pattern will become less and less sharp.

(ii) At the centre there will be white fringe followed by red colour

fringes on either side.

OR

R

½

½

½

½

½

½

1

1

5

(a) Expression for the Amplitude and the conditions 3

(b) Effect on Interference fringes 1 +1

(a) Diagram 1

Mathematical Proof 1 ½

Graph for δ 1

Conditions ½

(b) Relation to µ ½

Value of µ ½

of 16 Final draft 17/03/15 04:30p.m.

(a)

In the quadrilateral AQNR at Q and R , two of the angles are right

angles.

Therefore , the sum of the other angles of the quadrilateral is 1800

From the triangle QNR ,

Comparing these two equations

The total deviation δ is the sum of the deviations at the two faces

i.e.

will be minimum for

(b)

If A = 600

1

½

½

½

1

½

½

½

5

MARKING SCHEME SET 55/1/RU Q. No. Expected Answer ......MARKING SCHEME SET 55/1/RU Q. No. Expected...

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